Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (T a : v) os << a << " ";
  os << "}";
  return os;
}
namespace deb {
string arg_name;
void dump(int idx) { cerr << endl; }
template <class Head, class... Tail>
void dump(int idx, Head&& head, Tail&&... tail) {
  int jdx = idx;
  while (jdx < arg_name.length() && arg_name[jdx] != ',') jdx++;
  cerr << arg_name.substr(idx, jdx - idx) << "=" << head << ", ";
  dump(jdx + 2, forward<Tail>(tail)...);
}
}  // namespace deb
int main(void) {
  int n, q;
  double sy, a, b;
  scanf("%lf %lf %lf", &sy, &a, &b);
  scanf("%d", &n);
  vector<double> l(n + 2), r(n + 2), sum(n + 2);
  l[0] = r[0] = -1e9;
  l[n + 1] = r[n + 1] = 1e9 + 1;
  for (int i = 1; i <= n; i++) {
    scanf("%lf %lf", &l[i], &r[i]);
    sum[i] = sum[i - 1] + r[i] - l[i];
  }
  sy = -sy;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    double x, y, u, v, ans = 0;
    scanf("%lf %lf", &x, &y);
    u = a + (x - a) * sy / (y + sy);
    v = b + (x - b) * sy / (y + sy);
    int li = distance(r.begin(), upper_bound(r.begin(), r.end(), u));
    int ri = distance(l.begin(), lower_bound(l.begin(), l.end(), v)) - 1;
    if (l[li] < v) {
      ans += min(r[li], v) - max(l[li], u);
      if (li < ri) {
        ans += min(r[ri], v) - max(l[ri], u);
        ans += sum[ri - 1] - sum[li];
      }
    }
    printf("%.12lf\n", ans * (y + sy) / y);
  }
}